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User blog:Rgetar/Definition of standard form using iterated fundamental sequence
sto, nfs, sf, ots Let we have some ordinal notation sto ("sto" means "string to ordinal"), that is correspondence "string - ordinal": :α = sto(s), where s is string, α is ordinal. Let we have algorythm nfs ("nfs" means "notation fundamental sequence") of calculating element of fundamental sequence of ordinal, corresponding to a string in this notation: :αn = sto(nfs(s, k)), where s, k are strings; α, n are ordinals; α = sto(s), n = sto(k). Despite we have nfs, we yet have not system of fundamental sequences, since s and k may be not unique strings, corresponding to ordinals α and n. It may turn out that :s1 ≠ s2 or k1 ≠ k2 :sto(s1) = sto(s2) :sto(k1) = sto(k2) :sto(nfs(s1, k1)) ≠ sto(nfs(s2, k2)) that is we can get different fundamental sequences for the same ordinal. For each ordinal α this notation can work with we can choose one string, corresponding to α and name it "standard form", and all other strings in this notation, corresponding to α, we can name "non-standard forms". We can create algorythm sf ("sf" means "standard form"), converting strings into standard forms of corresponding ordinals in this notation: :α = sto(s) :α = sto(sf(s)) :sf(sf(s)) = sf(s) We can define function ots ("ots" means "ordinal to string"), calculating standard form of an ordinal in this notation: :ots(sto(s)) = sf(s) If we have nfs and sf, then we already have system of fundamental sequences. Standard form defined using nfs From my experience it turned out, that calculating of element of fundamental sequence (nfs) is easier than comparison of ordinals, corresponding to strings and converting strings into standard forms (sf); and comparison of ordinals, corresponding to strings, is a bit easier than converting strings into standard forms (sf). I made up an idea: define standard form using nfs, that is say: "If s and k are standard forms, then let nfs(s, k) is standard form by definition". The problem is that form nfs(s, k) may be not unique for ordinal sto(nfs(s, k)). It may turn out that :s1 ≠ s2 or k1 ≠ k2 :s1 = sf(s1) :s2 = sf(s2) :k1 = sf(k1) :k2 = sf(k2) :sto(nfs(s1, k1)) = sto(nfs(s2, k2)) :nfs(s1, k1) ≠ nfs(s2, k2) that is the same ordinal sto(nfs(s1, k1)) = sto(nfs(s2, k2)) may have different such forms: nfs(s1, k1) ≠ nfs(s2, k2). To solve this problem we can take some large ordinal α, choose a string s for it in this notation, say that s is standard form of α (that is ots(α) = s), and for each ordinal β < α take two ordinals γβ ≤ α and nβ ≤ α and say that ots(β) = nfs(ots(γβ), ots(nβ)). (That is for each β we choose one form nfs(sβ, kβ), where sβ = ots(γβ), kβ = ots(nβ)). We can express an ordinal ≤ α as αn0n1n2...nm with natural number of ni's and name it "iterated fundamental sequence form". Standard iterated fundamental sequence form To choose γβ and nβ for each β we can do the following. We start from α. We can say that α is "standard iterated fundamental sequence" (sifs) form of α. Then for each αn0 γαn0 is α, and nαn0 is n0. (That is all αn0 are sifs). We get sequence of ordinals :α0, α1, α2, α3, α4, α5, ... α and intervals between them :α[0) :(α0, α1) :(α1, α2) :(α2, α3) :(α3, α4) :(α4, α5) :... These intervals correspond to these ordinals (an ordinal corresponds to interval between supremum of part of this sequence consisting of lesser ordinals and this ordinal): :α[0) - α0 :(α0, α1) - α1 :(α1, α2) - α2 :(α2, α3) - α3 :(α3, α4) - α4 :(α4, α5) - α5 :... Then we choose αn0n1 such as they are within intervals, corresponding to αn0. (That is, we do not take, for example, α31, if it is within interval (α1, α2), but if α31 is within (α2, α3), then we take it). If αn0n1 is within interval, corresponding to αn0, then it is sifs. Otherwise, it is not sifs. We get another sequence, consisting of α, αn0 and sifs αn0n1, and intervals between them. Each αn0n1 also corresponds to an interval, and we choose αn0n1n2 such as they are within intervals, corresponding to αn0n1, and say that they are sifs, otherwise, it is not sifs. And so on. (That is we divide ordinals up to α into gaps with elements of funamental sequence of α, then we fill each gap with elements of funamental sequence of one ordinal, dividing them into sub-gaps, and so on, until we have no gaps left). Formally, let set S0 and set T0 contain one element α, that is S0 = T0 = {α}. Tk + 1 is set of all βn such as β ∈ Tk, βn > all γ such as γ ∈ Sk and γ < β. Sk + 1 = Sk ∪ Tk + 1. "α" is sifs. "βn" is sifs iff β ∈ Tk and βn ∈ Tk + 1. Possible problems here: I am not sure, do all β < α have sifs form (at least, with finite number of ni's). And, maybe, we need different α's for different cardinalities instead of one α. Cut systems of fundamental sequences Still we have a problem: we need to compare βn with elements of Sk to check, are they sifs, but comparison also may be not very easy. For each sifs β we have two possible cases: :1. All βn are sifs, or βn are partially sifs, and partially not sifs. :2. All βn are not sifs. If βn are partially sifs, and partially not sifs, then all βn, beginning from least sifs βn, are sifs, and all lesser βn are not sifs. If we have system of fundamental sequences and α, then we can cut this system of fundamental sequences, that is do such transformation: for case 1 cut each fundamental sequence, that is start it from least sifs element, and for case 2 do nothing. Or, we can make up a system of fundamental sequences, which is already cut, and prove that it is cut. For cut system of fundamental sequences we have two cases: :1. All βn are sifs. :2. All βn are not sifs. For cut system we do not need to compare. We only need to distinguish case 1 from case 2. I think that system of fundamental sequences from my square brackets notation may be cut, or, at least, can be remade into cut. I quess, that if β is successor, then it is case 2, and if β is limit, then it is case 1. (For β = 0 we have no elements of fundamental sequence). In this notation ordinal is successor, if it starts with "[]". Category:Blog posts